How to evaluate the infinite series: $ frac 1 {3cdot6} + frac 1 {3cdot6cdot9} +frac 1...
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The infinite series is given by:
$$ dfrac 1 {3cdot6} + dfrac 1 {3cdot6cdot9} +dfrac 1 {3cdot6cdot9cdot12}+ldots$$
What I thought of doing was to split the general term as:
$$begin{align}
t_r &= dfrac 1 {3^{r+1}(r+1)!}\\
&= dfrac {r+1 - r} {3^{r+1}(r+1)!}\\
&= dfrac {1} {3^{r+1}cdot r!} - dfrac{r}{3^{r+1}(r+1)!}
end{align}$$
But this doesn't seem to help.
HINTS?
calculus sequences-and-series limits
edited Nov 22 at 13:22
Martin Sleziak
44.4k7115268
asked Sep 2 '13 at 17:43
Parth Thakkar
2,73521635
add a comment |
up vote
2
down vote
favorite
The infinite series is given by:
$$ dfrac 1 {3cdot6} + dfrac 1 {3cdot6cdot9} +dfrac 1 {3cdot6cdot9cdot12}+ldots$$
What I thought of doing was to split the general term as:
$$begin{align}
t_r &= dfrac 1 {3^{r+1}(r+1)!}\\
&= dfrac {r+1 - r} {3^{r+1}(r+1)!}\\
&= dfrac {1} {3^{r+1}cdot r!} - dfrac{r}{3^{r+1}(r+1)!}
end{align}$$
But this doesn't seem to help.
HINTS?
calculus sequences-and-series limits
edited Nov 22 at 13:22
Martin Sleziak
44.4k7115268
asked Sep 2 '13 at 17:43
Parth Thakkar
2,73521635
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The infinite series is given by:
$$ dfrac 1 {3cdot6} + dfrac 1 {3cdot6cdot9} +dfrac 1 {3cdot6cdot9cdot12}+ldots$$
What I thought of doing was to split the general term as:
$$begin{align}
t_r &= dfrac 1 {3^{r+1}(r+1)!}\\
&= dfrac {r+1 - r} {3^{r+1}(r+1)!}\\
&= dfrac {1} {3^{r+1}cdot r!} - dfrac{r}{3^{r+1}(r+1)!}
end{align}$$
But this doesn't seem to help.
HINTS?
calculus sequences-and-series limits
edited Nov 22 at 13:22
Martin Sleziak
44.4k7115268
asked Sep 2 '13 at 17:43
Parth Thakkar
2,73521635
The infinite series is given by:
$$ dfrac 1 {3cdot6} + dfrac 1 {3cdot6cdot9} +dfrac 1 {3cdot6cdot9cdot12}+ldots$$
What I thought of doing was to split the general term as:
$$begin{align}
t_r &= dfrac 1 {3^{r+1}(r+1)!}\\
&= dfrac {r+1 - r} {3^{r+1}(r+1)!}\\
&= dfrac {1} {3^{r+1}cdot r!} - dfrac{r}{3^{r+1}(r+1)!}
end{align}$$
But this doesn't seem to help.
HINTS?
calculus sequences-and-series limits
calculus sequences-and-series limits
edited Nov 22 at 13:22
Martin Sleziak
44.4k7115268
asked Sep 2 '13 at 17:43
Parth Thakkar
2,73521635
edited Nov 22 at 13:22
Martin Sleziak
44.4k7115268
asked Sep 2 '13 at 17:43
Parth Thakkar
2,73521635
edited Nov 22 at 13:22
Martin Sleziak
44.4k7115268
edited Nov 22 at 13:22
Martin Sleziak
44.4k7115268
edited Nov 22 at 13:22
Martin Sleziak
44.4k7115268
44.4k7115268
asked Sep 2 '13 at 17:43
Parth Thakkar
2,73521635
asked Sep 2 '13 at 17:43
Parth Thakkar
2,73521635
asked Sep 2 '13 at 17:43
Parth Thakkar
2,73521635
2,73521635
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
7
down vote
accepted
HINT:
$$e^x=sum_{0le r<infty}frac{x^r}{r!}$$
Can you take it from here?
A strongly resembling sequence $$-ln(1-x)=sum_{1le r<infty}frac{x^r}{r}$$ for $-1le x<1$
answered Sep 2 '13 at 17:44
lab bhattacharjee
221k15154271
Sheesh! Didn't even think that it was $e^x$. I kept on trying to bring in $e$ somehow, but this, I couldn't think. Thanks :D
– Parth Thakkar
Sep 2 '13 at 17:46
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
HINT:
$$e^x=sum_{0le r<infty}frac{x^r}{r!}$$
Can you take it from here?
A strongly resembling sequence $$-ln(1-x)=sum_{1le r<infty}frac{x^r}{r}$$ for $-1le x<1$
answered Sep 2 '13 at 17:44
lab bhattacharjee
221k15154271
Sheesh! Didn't even think that it was $e^x$. I kept on trying to bring in $e$ somehow, but this, I couldn't think. Thanks :D
– Parth Thakkar
Sep 2 '13 at 17:46
add a comment |
up vote
7
down vote
accepted
HINT:
$$e^x=sum_{0le r<infty}frac{x^r}{r!}$$
Can you take it from here?
A strongly resembling sequence $$-ln(1-x)=sum_{1le r<infty}frac{x^r}{r}$$ for $-1le x<1$
answered Sep 2 '13 at 17:44
lab bhattacharjee
221k15154271
Sheesh! Didn't even think that it was $e^x$. I kept on trying to bring in $e$ somehow, but this, I couldn't think. Thanks :D
– Parth Thakkar
Sep 2 '13 at 17:46
add a comment |
up vote
7
down vote
accepted
up vote
7
down vote
accepted
HINT:
$$e^x=sum_{0le r<infty}frac{x^r}{r!}$$
Can you take it from here?
A strongly resembling sequence $$-ln(1-x)=sum_{1le r<infty}frac{x^r}{r}$$ for $-1le x<1$
answered Sep 2 '13 at 17:44
lab bhattacharjee
221k15154271
HINT:
$$e^x=sum_{0le r<infty}frac{x^r}{r!}$$
Can you take it from here?
A strongly resembling sequence $$-ln(1-x)=sum_{1le r<infty}frac{x^r}{r}$$ for $-1le x<1$
answered Sep 2 '13 at 17:44
lab bhattacharjee
221k15154271
edited Sep 2 '13 at 17:49
answered Sep 2 '13 at 17:44
lab bhattacharjee
221k15154271
answered Sep 2 '13 at 17:44
lab bhattacharjee
221k15154271
answered Sep 2 '13 at 17:44
lab bhattacharjee
221k15154271
221k15154271
Sheesh! Didn't even think that it was $e^x$. I kept on trying to bring in $e$ somehow, but this, I couldn't think. Thanks :D
– Parth Thakkar
Sep 2 '13 at 17:46
add a comment |
Sheesh! Didn't even think that it was $e^x$. I kept on trying to bring in $e$ somehow, but this, I couldn't think. Thanks :D
– Parth Thakkar
Sep 2 '13 at 17:46
Sheesh! Didn't even think that it was $e^x$. I kept on trying to bring in $e$ somehow, but this, I couldn't think. Thanks :D
– Parth Thakkar
Sep 2 '13 at 17:46
Sheesh! Didn't even think that it was $e^x$. I kept on trying to bring in $e$ somehow, but this, I couldn't think. Thanks :D
– Parth Thakkar
Sep 2 '13 at 17:46
add a comment |
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